Example

The loss of heat from a certain building, by conduction through the walls and windows, is 200,000 B. T. U. per hour. It is desired to heat the building by the indirect system with natural draft, with steam having a temperature of 220°. The average temperature of the hot air on entering the rooms should be 40° above that in the rooms. The building is two stories high, and all the radiators should be located in the basement. How many square feet of radiator surface will be required to maintain the internal temperature of the building (neglecting the basement and attic) at a temperature of 70°, with the outer air at zero?

Solution

The losses of heat are as follows:

By conduction, 40/110 of the total,

or

200,000

B.

T.

U.

By ventilation, 70/110 of the total,

or

350,000

B.

T.

U.

The total loss per hour is

550,000

B.

T.

U.

The height of the flues we will say is about ten feet for the first story, and 20 to 25 feet for the second story; the coefficients of emission from 2 + 2.5 the radiators will be 2+2.5 / 2= 2.25 B. T. U. per hour. The difference between the temperatures of steam and the cold outer air is 220°. Then the required area is 550,000 / 2.25 X 220 1,111 sq. ft. Ans.

92. Baldwin's Rule

Baldwin's Rule. One of the most simple, and probably most correct, empirical rules used for computing the size of direct radiators is that originated by Mr. Wm. J. Baldwin; it is as follows:

Rule 7

Divide the difference in temperature between that at which the room is to be kept and the coldest outside atmosphere, by the difference between the temperature of the steam pipes and that at which you wish to keep the room, and the quotient will be the square feet or fraction thereof of plate or pipe surface to each square foot of glass, or its equivalent in wall surface.

The quantity of heating surface found by this simple rule compensates only for the amount of heat lost by transmission through the windows, walls, and other cooling surfaces. It does not provide for cold air entering the room through loosely fitting doors, windows, etc., and an ample allowance must be made for this. Some buildings are so poorly constructed that 50 per cent. or more must be added to the amount of heating surface obtained by the above rule in order to counteract the cooling effect of these air leakages. A common practice is to add 25 per cent. for buildings of ordinary good construction. Besides this addition for air leakage, an ample allowance should be made for rooms exposed to cold winds, and this allowance should, if possible, be made in the form of an auxiliary radiator to prevent overheating the rooms during moderate weather.

93. As an example, suppose that we have three rooms A, B, -and C, as shown in Fig. 36, of precisely the same dimensions, and consequently having the same cubical contents, each room being 25 feet long by 20 feet wide, with a 10-foot ceiling. Let us also suppose that the halls, or corridors, D, and the other rooms in the building will be warmed to a temperature equal to that desired in A, B, and C by other radiators not shown. We are to find by the above rule the amount of heating surface required to maintain a temperature of 70° F. in A, B, and C, assuming that the radiators will be heated by steam having a pressure of 5 pounds by the gauge, when the outside temperature is 10° below zero. Let us suppose that the windows are each 6 ft. X 3 ft., and that the exposed walls are built of ordinary good brick, and that they are lathed and plastered inside.

92 Baldwin s Rule 145

Fig. 36.

Let S = amount of radiating surface required to counteract the cooling effect of the glass and its equivalent in exposed wall surface in square feet; t = difference in degrees F. between the desired temperature of the room and that of the external air; tl = difference in degrees F. between the temperature of the heating surface and that of the air in the room; s = number of square feet of glass and its equivalent in exposed wall surface.

Then, applying Baldwin's rule,

S

=

t

s.

t1

When lathed-and-plastered brick walls are used, as in the figure, it is safe to estimate that about 10 square feet of wall surface will be equivalent in cooling power to 1 square foot of glass; consequently, in this case, wall surface

=

equivalent glass surface.

10

Let us commence with the room A; the amount of glass surface here is 6 X 3 X 4 = 72 square feet. To this must be added the exposed wall surface reduced to a glass equivalent; thus,

10

(25 + 20)

-

72

=

37.8 sq. ft.

10

Since we assume that the inner walls, floors, and ceilings are not cooling surfaces, then the only cooling surfaces we have to allow for in the case of A is 72+37.8 = 109.8 square feet = s.

By substituting in the formula, we have, since the temperature of steam at 5 pounds gauge pressure is 227°, and the difference between 70° above zero and 10° below zero is 70° +10°,

S

=

70 + 10

X

109.8

=

56

sq. ft., nearly.

227 - 70

This, however, only counteracts the cooling effect of the walls and windows, and to make reasonable allowance for air leakage, we will add 25 per cent., or 14 square feet, which gives us 50 +14 = 70 square feet of direct radiating surface. Now, suppose that we allow 20 per cent. of the direct radiating surface (70 square feet in this case) for a moderate exposure to winds; the amount of heating surface, that is, the radiator which we would place in A, will have an area of 70 + 14, or 84 square feet.

For convenience, we may divide this into two radiators, a having an area of 56 square feet, and b an area of 28 square feet. This will so divide the radiator surface that one-third, or 28 square feet, may be used for duty during- mild weather; two-thirds, or 56 square feet, for moderate cold weather; and the whole, or 84 square feet, for use during severe weather.

In like manner and under the same conditions, we find that the sizes of the radiators c, d, and e should be, respectively, 40, 82, and 42 square feet.

As the coldest winds blow in the direction of the arrow, we place the 82-square-foot radiator in the left-hand exposed corner of the room C. A better distribution of the radiator surface in this room would be to make d 42 square feet only, and place a radiator having 40 square feet between the windows towards which the arrow points; this will give a more uniform temperature to the room.

94. It will be observed that A, B, and C, which are three rooms having the same shape and cubical contents, respectively require 84 square feet, 40 square feet, and 1.24 square feet of heating surface, in order to maintain a temperature of 70° F. in each while the outer atmosphere is 10° below zero. This shows how imperfect must be the rule-of-thumb method of proportioning radiators to the cubical contents of the several rooms.

For direct-indirect heating, the area of radiator surface required may be found by computing the area required for direct heating, and adding 25 per cent. to that amount. Thus, in the example in Art. 90, if the heating be done by the direct-indirect method instead of the direct method, the radiator surface required would be 762 + 190.5 = 952.5 square feet.